Question

Functions, Help?

Answer 1

(i) `x<3/2 or x>3`

(ii) `(3/4,-9/8)`

(iii) `k=27/8`

Explanation:

(i)

`y=2x^2-3x`

When `y>9`,

`2x^2-3x>9`

Subtract `9` from both sides,

`2x^2-3x-9>0`

Factor,

`(2x+3)(x-3)>0`

Graph,

graph{2x^2-3x-9 [-5, 5, -15, 15]}

From the graph,

`x<3/2 or x>3`

(ii)

`2x^2-3x`

Factor,

`2(x^2-3/2x)`

Perform complete the square method,

Let us recap our concepts,

`color(red)(a^2)+color(green)(2ab)+color(blue)(b^2)=color(purple)((a+b)^2`

Let `a=x` and `2ab=-3/2x`

`b=-(3/2x)/(2x)`
`color(white)(b)=-3/4`

Now we have to add in `color(blue)((-3/4)^2)` somehow,

`2(color(red)(x^2)-color(green)(3/2x)+color(blue)((-3/4)^2)-(-3/4)^2)`

By adding and subtracting, I have not affected the original equation,

`2(color(red)(x^2)-color(green)(3/2x)+cancel(color(blue)((-3/4)^2))-cancel((-3/4)^2))`
`=2(x^2-3/2x)`

Now simplify,

`2(color(red)(x^2)-color(green)(3/2x)+color(blue)((-3/4)^2))-2(-3/4)^2`

Make the perfect square and simplify,

`2color(purple)((x-3/4)^2)-9/8`

Hence,

`2(x-3/4)^2-9/8` - ( vertex form )

The vertex formula is as follows:

`a(x-h)^2+k`, where `(h,k)` is the vertex.

By comparison, find the coordinates,

`(3/4,-9/8)`

(iii)

`f(x)=2x^2-3x`

`g(x)=3x+k`

Substitute,

`gf(x)=g(2x^2-3x)`
`color(white)(gf(x))=3(2x^2-3x)+k`

When `gf(x)=0`,

`3(2x^2-3x)+k=0`

Expand,

`6x^2-9x+k=0`

Apply discriminant, where `b^2-4ac=0` for equal roots.

`(-9)^2-4(6)(k)=0`

Simplify,

`81-24k=0`

Add `24k` to both sides,

`24k=81`

Solve,

`k=81/24`
`color(white)(k)=27/8`